Question:
If
\( \sin^{-1}\left(\dfrac{2a}{1+a^2}\right) - \cos^{-1}\left(\dfrac{1-b^2}{1+b^2}\right) = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right) \),
then the value of \( x \) is:
If
\( \sin^{-1}\left(\dfrac{2a}{1+a^2}\right) - \cos^{-1}\left(\dfrac{1-b^2}{1+b^2}\right) = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right) \),
then the value of \( x \) is:
\( \sin^{-1}\left(\dfrac{2a}{1+a^2}\right) - \cos^{-1}\left(\dfrac{1-b^2}{1+b^2}\right) = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right) \),
then the value of \( x \) is:
Show Hint
Memorize inverse trigonometric standard forms.
Updated On: Mar 23, 2026
- \(\dfrac{a}{b}\)
- \(\dfrac{ab}{1+ab}\)
- \(\dfrac{b}{a}\)
- (a-b)/(1+ab)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use identities:
\( \sin^{-1}\left(\dfrac{2a}{1+a^2}\right) = 2\tan^{-1} a \),
\( \cos^{-1}\left(\dfrac{1-b^2}{1+b^2}\right) = 2\tan^{-1} b \)
Step 2:
\( 2(\tan^{-1} a - \tan^{-1} b) = 2\tan^{-1} x \)
Step 3:
\( x = \dfrac{a - b}{1 + ab} \)
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